I only estimated the compression of the rubber at 0.02", it could be off by a bit
In order for the buffer to just touch the rear of the receiver extension at zero velocity, it would have to have an initial velocity of about 4.5 fps max, because that is all the energy a standard spring can absorb.
But, lets work with the 11 fps you suggest.
Cycle time (from one primer ignition to the next primer ignition) is 0.08 seconds for a standard M4, as you can see in the graphs above. If we assume the hammer fall, and bolt return times remain the same [1], then the only difference will be the time from unlock to full rear. That can easily be calculated with the average velocity (initial velocity minus end velocity divided by two) and the distance traveled (3.75").
The standard M4 unlock to full rear time is 0.025 sec and the reduced is 0.057 seconds. The total cycle time for the reduced will be 0.112 seconds. The cyclic rate for the reduced would be about 500 rpm.
Extensive testing of the M16 and M4 has shown that if the cyclic rate drops below 650 rpm reliability tanks, and below 550 rpm it almost wonât function at all. Service weapons are required to have 700 to 970 rpm.
Why is this the case?
Extraction.
In an AK or AR-18 the gas piston is still under pressure and pushing the bolt back as extraction takes place. With the AR-15 bolt design, extraction starts only after all forces acting on the bolt carrier cease, leaving only the momentum of the bolt carrier to provide all the energy for extraction. That energy is tied to the carrier velocity squared, so small reductions in velocity make large reductions in energy.
- It will actually be a bit slower as the elasticity of the rubber bumper will return an little energy to the forward push, but weâll ignore that for now.
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m = mass of BCG+buffer = 14oz
V0 = bolt initial speed = 15fps (fig. 2)
V1 = bolt speed when buffer hits tube end = 9fps (fig.2)
Energy stored in spring = 1/2m(V0^2-V1^2). In order for the buffer not contacting the tube end, the bolt should have initial speed V0â given by
1/2mV0â^2=1/2m(V0^2-V1^2)
V0â=sqrt(V0^2-V1^2)=12fps
The energy stored in the spring is about 2 ft-lb. As long as this is satisfied, the cycling should be no problem, regardless the cycling rate. But there is no margin should situation happens, cold climate for instance. It makes sense to crank up gas a bit more.
With adjustable GB, I do this all the time. Open the valve till the the bolt just holds open on last round. Then open a little more. At that the buffer tip would barely contact the buffer tube rear end.
-TL
You still have the extraction energy problem. Once extraction has started, there are no more energy inputs to the system.
Iâll wager your buffer is hitting the back of the extension harder than you think it is.
Your graphs have already included every steps of the cycling, do they not? Extraction, ejection, feeding, chambering, locking. No additional energy will be needed. They must as it was data from firing bursts. I got the numbers from there.
The buffer may be contacting the tube end, but barely. Certainly I canât stick my finger in there to feel it. But I can tell from the felt recoil pattern. Over gasing is boinâŚWRAMP. Under gasing (no last round lock back) is boin⌠I set for boin⌠tiny wramp.
Ah come on now. This thing canât be that new.
-TL
For the OPâŚ
The âpounds-feetâ measurement is misleading.
This site uses a more standard/relevant measurement in simple âpounds.â
(Iâve seen similar figureds in Shooting Times, which seems to be the only gun rag that gets tech right.)
The graphs show velocity and displacement, not energy. You can see the effect of extraction on the velocity, note that the velocity graphs are flatter on top for the lower velocity bolts.
The flat top gets wider as the initial velocity drops off and the steeper the velocity curve. I have never seen a trace like this where the bolt travels back far enough to reliably lock open that did not contact the back of the extension.
In contrast, you have the H&K 416, because the piston provides energy for extraction, you can get a lower bolt velocity and maintain reliable extraction. Note there is no flat spot on the top of the velocity curve.
Compare that to the M4 curve:
Your buffer is making contact with the receiver extension. The velocity, however is very low, and the force generated by the impact is close to imperceptible.
Kinetic energy is function of speed. E=1/2mV^2
Any positive jump speed on that graph means the bolt has energy to spare after completing extraction and ejection. Part of the spared energy is stored in the spring for feeding and chamber down the road. The jump to negative is caused by the buffer hitting the tube end hard and bouncing back.
Iâm not familiar with the HKâs construction. I believe it doesnât have a buffer but a very stout spring. It decelerates the bolt without going to coil bind. So there is no jump back to negative.
The buffer hits the tube end with very little force. Thatâs the point Iâm trying to make. You donât need a fancy buffer to do the same.
-TL
Lysander, you supply wonderful data, as usual. And itâs nice to see some basic classical mechanics applied to our favorite machine.
In your second set of graphs (the M4 set), I see a bcg with a rearward velocity of maybe 9fps hitting the end of the buffer tube and very quickly getting a forward (negative) velocity of around 4 fps. This doesnât seem like a minor collision to me.
By the way,these figures agree with some experiments Iâve done, which suggest that the pliable buffer tip absorbs about 4/5 of the energy of the buffer/buffer tube collision.
Speaking of which, remember that the buffer and bcg is not one piece, and separation and recontact can occur. This must occur during extraction, and may occur during the initial hammer blow of the high pressure gas first reaching the expansion chamber.
You mention an extraction effect on velocity in your graphs, and Iâm not seeing it.
Compare the velocity curve at the start of the cycle and compare it to the same place on the M27 curve that has a slight upward curve.
And there has to be energy loss during extraction, because work is being done (unsticking, and accelerating the empty case) by a system that no longer has any energy inputs.
The data sets are quite noisy. I see some sharp velocity drops of different extents in the transition area between acceleration and deceleration, sometimes multiple ones. These seem to range between 1/10 and 1/5 of the total kinetic energy, which seems reasonable for extraction energy.
But on some cycles there are multiple sharp drops, and sharp rises too of roughly the same amplitude.
Some of this may be due to the bcg body colliding with the bolt as the cam pin reaches the end of the track.
Maybe when I look at the plots with smoothing eyes I can see a not-very-consistent extraction effect. Of course we do expect variation across cycles, especially in extraction energy.
Can you supply a link to the source paper?
And the collision with the end of the buffer tube is of a smaller magnitude than I first thought, about like the bcg/buffer hitting a table after falling about 1-1/4 feet (since v=(2 x g x falling distance)^(1/2)).
They come from several reports, I canât remember which come from which, but I think it might be from these:
AD B219959 - Abbreviated Report for the Initial Production Test (IPT) of the Carbine System, 5.56mm, M4A1
AD B402939 - M4 Reliability Analysis
AD B324971 - System Assessment For PEO Soldier Baseline Reliability And Dust Assessment For The M4, M16, And M249
AD 1013851 - 2015 Re-Baseline Reliability Test 5.56mm Weapons Using M855A1
All of that is voodooâŚ
Shrink your gas port and a plain old Carbine buffer is perfect for semi-auto and an H is good for full-auto.